The maximum and minimum values of cos^{6} the | vedclass

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(A) Let $f(	heta) = \sin^{6} 	heta + \cos^{6} 	heta$.Using the identity $a^{3} + b^{3} = (a + b)(a^{2} - ab + b^{2})$,we have:$f(	heta) = (\sin^{2

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$1$ and $\frac{1}{4}$

Vedclass · Answer

(A) Let $f(	heta) = \sin^{6} 	heta + \cos^{6} 	heta$.Using the identity $a^{3} + b^{3} = (a + b)(a^{2} - ab + b^{2})$,we have:$f(	heta) = (\sin^{2} 	heta + \cos^{2} 	heta)(\sin^{4} 	heta - \sin^{2} 	heta \cos^{2} 	heta + \cos^{4} 	heta)$.Since $\sin^{2} 	heta + \cos^{2} 	heta = 1$,this simplifies to:$f(	heta) = (\sin^{2} 	heta + \cos^{2} 	heta)^{2} - 3 \sin^{2} 	heta \cos^{2} 	heta$.$f(	heta) = 1 - 3(\sin 	heta \cos 	heta)^{2}$.Using $\sin 2	heta = 2 \sin 	heta \cos 	heta$,we get $\sin 	heta \cos 	heta = \frac{1}{2} \sin 2	heta$.$f(	heta) = 1 - 3 \left(\frac{1}{2} \sin 2	hetaight)^{2} = 1 - \frac{3}{4} \sin^{2} 2	heta$.Since $0 \leq \sin^{2} 2	heta \leq 1$,the range of $f(	heta)$ is:For $\sin^{2} 2	heta = 0$,$f(	heta) = 1 - 0 = 1$ (Maximum).For $\sin^{2} 2	heta = 1$,$f(	heta) = 1 - \frac{3}{4} = \frac{1}{4}$ (Minimum).Thus,the maximum and minimum values are $1$ and $\frac{1}{4}$ respectively.

The maximum and minimum values of $\cos^{6} \theta + \sin^{6} \theta$ are respectively

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